How do you find the definite integral for: #[(6x^2+2) / sqrt(x)] dx # for the intervals #[1, 5]#?

2 Answers
May 23, 2017

#~~136.71#

Explanation:

#"rewrite as"#

#(6x^2)/(x^(1/2))+2/x^(1/2)=6x^(3/2)+2x^(-1/2)#

#rArrint_1^5(6x^(3/2)+2x^(-1/2))dx#

#"integrate each term using the "color(blue)"power rule"#

#• int(ax^n)=a/(n+1)x^(n+1); n!=-1#

#=[12/5x^(5/2)+4x^(1/2)]_1^5#

#=(12/5. 5^(5/2)+4. 5^(1/2))-(12/5 .1+4.1)#

#=60sqrt5+4sqrt5-32/5#

#=64sqrt5-32/5~~136.71" to 2 dec. places"#

May 23, 2017

# 64sqrt5-6.4#

Explanation:

Let, #I=int_1^5 (6x^2+2)/sqrtx dx.#

We know that, #intx^ndx=x^(n+1)/(n+1), n!=-1.#

#:. I=int_1^5(6x^2+2)/sqrtxdx,#

#=int_1^5(6x^2+2)*x^(-1/2)dx,#

#=int_1^5{(6x^(2-1/2)+2x^(-1/2)}dx,#

#=6int_1^5x^(3/2)dx+2int_1^5x^(-1/2)dx,#

#=6[x^(3/2+1)/(3/2+1)]_1^5+2[x^(-1/2+1)/(-1/2+1)]_1^5,#

#=6*2/5[x^(5/2)]_1^5+2*2[x^(1/2)]_1^5,#

#=12/5[5^(5/2)-1^(5/2)]+4[5^(1/2)-1^(1/2)]#

#=12/5(25sqrt5-1)+4(sqrt5-1),#

#=60sqrt5-12/5+4sqrt5-4,#

# rArr I=64sqrt5-6.4#

Enjoy Maths.!